Area of triangle

Chapter 7 Class 10 Coordinate Geometry
Concept wise

Introducing your new favourite teacher - Teachoo Black, at only ₹83 per month

### Transcript

Example 11 Find the area of a triangle whose vertices are (1, –1), (– 4, 6) and (–3, –5). Let the two points be A (1, −1) , B(−4, 6) & C(−3, −5) Area of triangle ABC = 1/2 [ x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2) ] Here x1 = 1 , y1 = −1 x2 = −4 , y2 = 6 x3 = −3 , y3 = −5 Putting values Area of triangle ABC = 1/2 [ 1(6 – (−5)) + (−4)(−5 – (−1) ) + (−3)(−1 − 6) ] Area of triangle ABC = 1/2 [1(6 – (−5)) + (−4)(−5 – (−1) ) + (−3)(−1 − 6)] = 1/2 [1(6 + 5) + (−4)(−5 + 1 ) + (−3)(−7)] = 1/2 [ 1(11) + (−4)(−4 ) + (−3)(−7) ] = 1/2 [ 11 + 16 + 21 ] = 1/2 [ 48 ] = 24 square units

Made by

#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.